View source for (c-a-b)2F1-(c-a)2F1(a-1)+b(1-z)2F1(b+1)=0

==Theorem==
The following formula holds:
$$(c-a-b){}_2F_1(a,b;c;z)-(c-a){}_2F_1(a-1,b;c;z)+b(1-z){}_2F_1(a,b+1;c;z)=0,$$
where ${}_2F_1$ denotes [[hypergeometric 2F1]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=(c-a-1)2F1+a2F1(a+1)-(c-1)2F1(c-1)=0|next=}}: $\S 2.8 (36)$

[[Category:Theorem]]
[[Category:Unproven]]